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\begin{document}

\title{高等代数一}
\subtitle{15-向量空间的基本性质}
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2022年11月10日} }

\maketitle

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\begin{frame}{内容提要：向量空间的基本性质}

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\begin{itemize}
\item[P1.]  在每个向量空间中，零向量是唯一存在的。
\item[P2.]  向量空间中的每个向量的负向量是唯一存在的。
\item[P3.]  数量零乘以任意向量的结果都是零向量。
\item[P4.]  任意数量乘以零向量的结果都是零向量。
\item[P5.]  若一个数乘运算的结果为零向量，则要么数量为零，要么向量为零向量。
\item[P6.]  数 $-1$ 乘以一个向量正好等于这个向量的负向量。
\end{itemize}


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\begin{itemize}
\item[P1.]  在每个向量空间中，零向量是唯一存在的。
\item 证明：
\begin{enumerate}
\item 设 $\theta_1$ 和 $\theta_2$ 都是向量空间 $V$ 的零向量。
\item 使用公理，证 $\theta_1=\theta_2$. 
\item $\theta_1 = \theta_2 + \theta_1 = \theta_1 + \theta_2 = \theta_2$.
\end{enumerate}

\end{itemize}

\vfill 

{\color{red}在本课程中，我们总是用小写希腊字母 $\theta$ 表示抽象的向量空间中的零向量。}

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\begin{itemize}
\item[P2.]  向量空间中的每个向量的负向量是唯一存在的。
\item 证明：
\begin{enumerate}
\item 设 $\beta_1$ 和 $\beta_2$ 都是向量 $\alpha$ 的负向量，即有 $\beta_1+\alpha =\theta$, $\beta_2+\alpha =\theta$. 
\item 使用公理A3, A2, A1 可得 
{\footnotesize 
\begin{eqnarray*}
\beta_1 &=& \theta+\beta_1 \\ 
&=& (\beta_2+\alpha)+\beta_1 \\  
&=& \beta_2+(\alpha+\beta_1) \\ 
&=& \beta_2+(\beta_1+\alpha) \\ 
&=& \beta_2+\theta \\ 
&=& \theta+\beta_2 \\ 
&=& \beta_2. 
\end{eqnarray*}
}



\end{enumerate}

\end{itemize}

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\begin{itemize}
\item[P3.]  数量零乘以任意向量的结果都是零向量。
\item 证明：
\begin{enumerate}
\item 记 $\beta = 0\cdot \alpha$. 
\item 由公理 A6 可证 $\beta+\beta = \beta$. 
\item 由公理 A4 和 A2 可证 $\beta =\theta$. 
\end{enumerate}

\end{itemize}

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\begin{itemize}
\item[P4.]  任意数量乘以零向量的结果都是零向量。
\item 证明：
\begin{enumerate}
\item 记 $\beta = k\cdot \theta$. 
\item 由公理 A5 可得 $\beta+\beta = \beta$. 
\item 证 $\beta =\theta$. 
\end{enumerate}

\end{itemize}

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\begin{itemize}
\item[P5.]  若一个数乘运算的结果为零向量，则要么数为零，要么向量为零向量。
\item 证明：
\begin{enumerate}
\item 设 $k\cdot \alpha = \theta$. 
\item 设 $k\neq 0$, 则 $\frac{1}{k} \in F$. 从而 $\frac{1}{k}\cdot (k\cdot \alpha)=\frac{1}{k}\cdot \theta$.  
\item 上式左边由公理 A7 和 A8 可得 $\alpha$, 上式右边由性质 P4 可得零向量。 
\item 从而得证 $\alpha=\theta$. 
\end{enumerate}

\end{itemize}

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\begin{itemize}
\item[P6.]  数 $-1$ 乘以一个向量正好等于这个向量的负向量。
\item 证明：
\begin{enumerate}
\item 要证的是 $(-1)\cdot \alpha = -\alpha$. 
\item 按负向量的定义（公理 A4），这是要证 $(-1)\cdot \alpha + \alpha = \theta$. 
\item 考虑公理 A8 和 A6, 再考虑性质 P3. 
\end{enumerate}

\end{itemize}

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\begin{itemize}

\item  例子1：如果 {\footnotesize $a(1,2,3)+b(2,3,1)+c(3,1,2)=(0,0,0)$}, 那么 {\footnotesize $a=b=c=0$}. 

\item  证明：由向量空间 $\mathbb{R}^3$ 中的向量的数乘与加法的定义，可得
{\footnotesize 
\begin{eqnarray*}
(a+2b+3c, 2a+3b+c, 3a+b+2c) = (0,0,0).
\end{eqnarray*}
}
根据{\color{red}两个向量相等的定义是分量对应相等}，可得 
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
a+2b+3c &=& 0, \\ 
2a+3b+c &=& 0, \\
3a+b+2c &=& 0.
\end{array}\right. 
\end{eqnarray*}
}
计算系数行列式的值为 $-18$, 不为零，所以该线性方程组只有零解。

\end{itemize}

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\begin{itemize}

\item  例子2：找出3个不全为零的实数 $a,b,c$ 使得 
{\footnotesize 
\begin{eqnarray*}
a(1,2,2)+b(2,1,3)+c(2,-5,1)=(0,0,0). 
\end{eqnarray*}
}
\item  解答：这个向量方程等价于下述线性方程组
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
a+2b+2c &=& 0, \\ 
2a+b-5c &=& 0, \\
2a+3b+c &=& 0.
\end{array}\right. 
\end{eqnarray*}
}
将系数矩阵化为{\color{red}行最简形}，可得通解 $(a,b,c)=k(4,-3,1)$, 其中 $k\in\mathbb{R}$.  
{\footnotesize 
\begin{eqnarray*}
\begin{pmatrix} 1&2&2 \\ 2&1&-5 \\ 2&3&1 \end{pmatrix} 
\to
\begin{pmatrix} 1&0&-4 \\ 0&1&3 \\ 0&0&0 \end{pmatrix}. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  例子3：设有数域 $F$ 上的向量空间 $V$, 设 $k\in F, \alpha,\beta\in V$, 证明 $$k(\alpha-\beta)=k\alpha-k\beta. $$

\item  证明：根据公理 A5, A7 与性质P6, 以及{\color{red}减法是加上负向量}，可得
{\footnotesize 
\begin{eqnarray*}
k(\alpha - \beta) 
&=& k(\alpha + (-\beta)) \\ 
&=& k\alpha + k(-\beta) \\ 
&=& k\alpha + k((-1)\beta) \\ 
&=& k\alpha + (-1)(k\beta) \\ 
&=& k\alpha + (-(k\beta))  \\ 
&=& k\alpha - k\beta. 
\end{eqnarray*}
}

\end{itemize}

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\begin{itemize}

\item  例子4：设有数域 $F$ 上的向量空间 $V$, 设 $V$ 中存在非零向量，即存在 $\alpha\in V$ 且 $\alpha\neq \theta$.  证明 $V$ 中有无限多个向量。

\item 证明：
\begin{enumerate}
\item  可证向量 $\alpha, 2\alpha, 3\alpha, \cdots $ 互不相同。
\item  设有正整数 $n,m$ 使得 $n\alpha=m\alpha$. 
\item  从 $n\alpha=m\alpha$ 可得 $(n-m)\alpha=\theta$. 
\item  根据性质 P5 可得 $n=m$. 
\end{enumerate}

\item  注：只有零向量的集合 $V=\{\theta\}$ 也是一个向量空间。

\end{itemize}

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\begin{enumerate}

\item  设有数域 $F$ 上的向量空间 $V$, 设 $\alpha\in V$, 证明 
{\footnotesize 
\begin{eqnarray*}
2\alpha &=& \alpha+\alpha, \\ 
3\alpha &=& \alpha+\alpha+\alpha. 
\end{eqnarray*}
}

\item  设有数域 $F$ 上的向量空间 $V$, 设 $k,m\in F, \alpha\in V$, 证明 $$(k-m)\alpha=k\alpha-m\alpha. $$


\end{enumerate}

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\begin{enumerate}

\item  解答思路：考虑公理 A8 与 A6. 
\item  解答思路：减去一个向量定义为加上这个向量的负向量。注意 $-m\alpha$ 是向量 $m\alpha$ 的负向量，向量 $m\alpha$ 是数量 $m$ 与向量 $\alpha$ 进行数乘的结果。
\end{enumerate}

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